Acceleration

• What acceleration means physically and mathematically
• The three equations of motion and when to use each
• How to handle deceleration (negative acceleration)
• Step-by-step application to distance, time, and velocity problems

What is acceleration? Acceleration is the rate of change of velocity with respect to time. If an object's velocity changes — whether it speeds up, slows down, or changes direction — it is accelerating.

a = (v - u) / t

where:

• a = acceleration (m/s²)
• v = final velocity (m/s)
• u = initial velocity (m/s)
• t = time taken (s)

Acceleration is a vector quantity — it has both magnitude and direction. If acceleration is in the same direction as motion, the object speeds up. If acceleration is opposite to motion, the object slows down (this is deceleration, or negative acceleration).

Uniform vs non-uniform acceleration Uniform acceleration: the velocity changes by the same amount every second (constant a). A freely falling object near Earth's surface has uniform acceleration of 9.8 m/s² downward.

Non-uniform acceleration: the velocity changes by different amounts in equal time intervals. A car in city traffic accelerates and decelerates irregularly — this is non-uniform.

Three equations of motion (for uniform acceleration only)

These three equations relate the five quantities: u (initial velocity), v (final velocity), a (acceleration), t (time), and s (displacement).

Equation 1: v = u + at Use this when you know u, a, t and want v.

Equation 2: s = ut + (1/2)at² Use this when you know u, a, t and want displacement s.

Equation 3: v² = u² + 2as Use this when time t is not given or not needed — relates v, u, a, and s directly.

Deceleration Deceleration is acceleration in the direction opposite to motion. It has a negative value when the positive direction is defined as the direction of motion. For example, if a car moving at 20 m/s brakes to rest in 4 s:

a = (0 - 20) / 4 = -5 m/s²

The negative sign means the acceleration opposes the velocity. In magnitude it is 5 m/s².

Graphical interpretation On a velocity–time graph, acceleration equals the slope (gradient) of the line. A steep upward slope = large positive acceleration; a downward slope = deceleration; a horizontal line = zero acceleration (constant velocity).

Problem: A car starts from rest and accelerates uniformly at 3 m/s² for 5 seconds. Find: (a) The final velocity. (b) The distance covered.

Solution:

Given: u = 0 m/s, a = 3 m/s², t = 5 s

(a) Final velocity — use Equation 1: v = u + at = 0 + (3)(5) = 15 m/s

(b) Distance covered — use Equation 2: s = ut + (1/2)at² s = (0)(5) + (1/2)(3)(5²) s = 0 + (1/2)(3)(25) s = 37.5 m

Verification using Equation 3: v² = u² + 2as 15² = 0 + 2(3)(s) 225 = 6s s = 37.5 m ✓

• Wrong sign for deceleration. If the positive direction is forward and the object is braking, acceleration must be negative: a = -|a|. Using a positive value gives wrong answers for both final velocity and distance.
• Applying equations of motion to non-uniform acceleration. The three equations only work when acceleration is constant. For variable acceleration, calculus or graphs are needed.
• Forgetting units for acceleration. Speed is m/s, so acceleration is m/s² (metres per second per second). Writing m/s for acceleration loses marks.
• Confusing s (displacement) with total path length. If an object reverses direction, displacement and distance differ. The equations give displacement, not total distance.

1. A ball is thrown upward at 20 m/s. Gravity decelerates it at 10 m/s². How long does it take to reach the highest point?
2. A train decelerates from 30 m/s to rest in 15 s. What is its acceleration?
3. Using v² = u² + 2as, find the distance a cyclist covers while accelerating from 2 m/s to 6 m/s at 2 m/s².